Question

For the principal value ${\sin }^{-1}\left(\sin 5\right)+{\cos }^{-1}\left(\cos 6\right)+{\tan }^{-1}\left(\tan 7\right)$ is equal to?

• A. $6\pi -18$
• B. $18-6\pi$
• C. $6-2\pi$
• D. $2\pi -8$

Answer 1

The correct option is C $6-2\pi$

here $5,6\in (\frac{3\pi }{2},2\pi )$ & $7\in (2\pi ,\frac{5\pi }{2})$

For principle value, we are given,

$-\frac{3\pi }{2}<5<2\pi$

$-\frac{3\pi }{2}>-5>-2\pi$

$2\pi -\frac{3\pi }{2}>2\pi -5>2\pi -2\pi$

$0<(2\pi -5)<\pi /2\to \left(1\right)$

so, $0<(2\pi -6)<\pi /2\to \left(2\right)$

$2\pi <7<\frac{5\pi }{2}$

$-2\pi >-7>\frac{-5\pi }{2}$

$2\pi -2\pi >2\pi -7>2\pi -\frac{5\pi }{2}$

$0>2\pi -7>-\pi /2.$

so, $-\pi /2<(2\pi -7)<0.\to \left(3\right)$

here $\sin 5=-\sin (2\pi -5),$ [Because of Fourth quadrant]

$\cos 6=\cos (2\pi -6)$ &

$\tan 7=-\tan (2\pi -7)$

so, ${\sin }^{-1}\left(\sin \right(5)+{\cos }^{-1}(\cos 6)+{\tan }^{-1}(\tan 7\left)\right)$

$={\sin }^{-1}(-\sin (2\pi -5)+{\cos }^{-1}(\cos (2\pi -6))+{\tan }^{-1}(\tan (2\pi -7))$

$=-(2\pi -5)+(2\pi -6)+(-1)(2\pi -7)$

$=-2\pi +5+2\pi -6-2\pi +7$

$=6-2\pi$